Question

2+4+6+8+10..........+92=

Answer 1

Hi ,

2 + 4 + 6 + ....+ 92

here ,

the terms are in A.P

first term = a = 2

common difference = d = a3 - a2

d = 6 - 4 = 2

last term = l = 92

let number of terms given = n

nth term = a + ( n - 1 ) d = 92

2 + ( n - 1 ) 2 = 92

( n- 1 ) 2 = 92 - 2

n-1 = 90 /2

n - 1 = 45

n = 45 + 1

n = 46

Therefore ,

sum of n terms = n /2 [ a + l ]

S45 = 46 /2 ( 2 + 92 )

= 23 × 94

=2162

I hope this helps you.

:)

Answer 2

4-2 = 2 
8-6 = 2 
since the difference between two consecutive terms is same therefore it forms an A.P.
⇒ first term , a = 2
    common difference , d = 2 
    last term , $a_{n}$  = 92 
By using the formula : $a_{n}$ = a + (n-1)d {n = no. of terms}
⇒ 92 = 2 + (n - 1)2
⇒ 92 - 2 = (n - 1)2
⇒ 90 = (n - 1)2
⇒ 90/2 = (n - 1)
⇒ 45 = (n - 1)
⇒ 45 + 1 = n
⇒ n =46 
now sum of n terms of an A.P is given by : $S_{n}$ = $\frac{n}{2}$(2a + (n - 1)d)
⇒ sum = $\frac{46}{2}$(2*2 + ( 46 - 1 )2)
⇒         = 23( 4 + 45*2)
⇒         = 23 ( 4 + 90)
⇒         = 23 (94)
⇒         = 2162 = answer